Find all 3 digit numbers such that by adding the first digit, the
square of the second and the cube of the third you get the original
number, for example 1 + 3^2 + 5^3 = 135.

In Between

Stage: 5 Challenge Level:

There were many correct solutions to this
problem: Alex from Stoke on Trent Sixth Form; Manuele and Eduardo,
both from the British School, Manila; Jeff from Sierra Vista;
Dapeng from Claremont Fancourt School, Diarmuid from Trinity
School, Nottingham; Christina from Smithdon High School; Thomas
from A Y Jackson School; Tathagata from South Point High School;
Jack from Guthlaxton College; Innyeong Chang from Nanjing
International School; Chen Jinsheng from ACS Independent School and
John from Takapuna Grammar School.

This is Alex's solution:

Multiply each side of the inequality by $\sqrt x$. There is no
problem about reversing the inequality sign, because $\sqrt x$ is
given as positive. This gives: $x + 1 < 4\sqrt x$.
Subtract $4\sqrt x$ from both sides so the right-hand side is
zero to give the following inequality involving a
quadratic in $\sqrt x$:
$$x - 4\sqrt x + 1 < 0.$$
Use the quadratic formula to factorise the left-hand side:
$$(\sqrt x - 2 -\sqrt 3)(\sqrt x - 2 + \sqrt 3) < 0.$$
Considering the value of the left-hand side for the positive values
of $\sqrt x$, the left-hand side is less than zero as required only
for the range
$$2 - \sqrt 3 < \sqrt x < 2 + \sqrt 3.$$
Here $x$ and $\sqrt x$ are both positive, and each value of $\sqrt
x$ corresponds to a sole $x$ value and vice-versa, so everything in
the inequality can be squared without losing solutions. The
solution of the inequality is:
$$7-4\sqrt 3 < x < 7+4\sqrt 3.$$
Note that $7-4\sqrt 3\approx 0.0718 > 0$.

Christina's method was slightly more
direct:

Square both sides. Multiply throughout by x. Rearrange to form the
quadratic inequality:
$$x^2-14x+1< 0.$$
Use the quadratic formula to solve this inequality. From the graph
of $y =x^2 -14x + 1$ we see that the solution is $7-4\sqrt 3 < x
< 7+4\sqrt 3$ or approximately $0.072< x< 13.928.$

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